Below is given a proof of a result. What result is provedProof Let a 2 (mod 4) and b 1
Chapter 5, Problem 5.66(choose chapter or problem)
Below is given a proof of a result. What result is proved?Proof Let a 2 (mod 4) and b 1 (mod 4) and assume, to the contrary, that 4 | (a2 + 2b). Sincea 2 (mod 4) and b 1 (mod 4), it follows that a = 4r + 2 and b = 4s + 1, where r,s Z. Therefore,a2 + 2b = (4r + 2)2 + 2(4s + 1) = (16r 2 + 16r + 4) + (8s + 2)= 16r 2 + 16r + 8s + 6.Since 4 | (a2 + 2b), we have a2 + 2b = 4t, where t Z. So 16r 2 + 16r + 8s + 6 = 4t and6 = 4t 16r 2 16r 8s = 4(t 4r 2 4r 2s).Since t 4r 2 4r 2s is an integer, 4 | 6, which is a contradiction.
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